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3 years ago

10

Someone please help I really appreciate it

2 answers:

Alex_Xolod [135]3 years ago

5 0

So you have to find the relationship between the pumpkin diameter (y) and the number of weeks passed (x) in a table, lets do it taking into account that the equation modeling such behaviour is:
y = 2x + 6, where x is the number of weeks and plus original 6 cm
x y
week diameter
0 6
1 8
3 12
5 16
10 26
substitute the x and y values in the equation to see how they fit into it

ss7ja [257]3 years ago

5 0

On the left column it would be 0,1,3,5,10. The right column would be 6,8,12,16,26

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Convert 2000 ft./min. to miles per hour round to the nearest 10th

Serggg [28]

Answer:

The conversion of 2000 ft per min into miles per hour is 2.268 miles per hour .

Step-by-step explanation:

Given as :

The speed = s = 2000 ft per min

Now, ∵ 5280 feet = 1 miles

so, 1 feet = $\dfrac{1}{5280}$ miles

∴ 2000 feet = $\dfrac{1}{5280}$ × 2000 miles

i.e 2000 feet = 0.378 miles

<u>Again</u>

∵ 1 minute = $\dfrac{1}{60}$ hour

So, The speed in miles per hour

s = $\dfrac{distance}{time}$

∴ s = $\frac{0.378}{\frac{1}{60}}$

i.e s = 2.268 miles per hour

Hence, The conversion of 2000 ft per min into miles per hour is 2.268 miles per hour . answer

8 0

3 years ago

The price of an article on sale was 70% of the regular price. If the sale price was $28, what was the regular price?

Doss [256]

Salei is 70%
28=70%
divide both sides by 7
4=10%
times both sides by 10
40=100%
regular price=$40

7 0

3 years ago

In algebra, we use an(n)___ to represent a variable.

Tamiku [17]

Answer:

A letter.

You can use any letter of the alphabet, but the ones that are most commonly used are x, y, and z.

7 0

3 years ago

The weights of steers in a herd are distributed normally. The standard deviation is 300lbs and the mean steer weight is 1100lbs.

gizmo_the_mogwai [7]

Answer:

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

<em> P(920≤ x≤1730) = 0.7078 </em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given mean of the Population = 1100 lbs

Standard deviation of the Population = 300 lbs

Let 'X' be the random variable in Normal distribution

Let x₁ = 920

$Z = \frac{x-mean}{S.D} = \frac{920-1100}{300} = - 0.6$

Let x₂ = 1730

$Z = \frac{x-mean}{S.D} = \frac{1730-1100}{300} = 2.1$

<u><em>Step(ii)</em></u>

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(x₁≤ x≤x₂) = P(Z₁≤ Z≤ Z₂)

= P(-0.6 ≤Z≤2.1)

= P(Z≤2.1) - P(Z≤-0.6)

= 0.5 + A(2.1) - (0.5 - A(-0.6)

= A(2.1) +A(0.6) (∵A(-0.6) = A(0.6)

= 0.4821 + 0.2257

= 0.7078

<u><em>Conclusion:-</em></u>

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

<em> P(920≤ x≤1730) = 0.7078 </em>

5 0

3 years ago

At the school cafeteria the ratio of pints of chocolate milk to pints of milk sold is 4:7. At this rate how mnay pints of milk w

pav-90 [236]

4:7
4*25:7*25
100:175
175 pints of milk

7 0

3 years ago

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